水题:UVa133-The Dole Queue
2017-08-29 17:08
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The Dole Queue
Time limit 3000 msDescription
In a serious attempt to downsize (reduce) the dole queue, The New National Green Labour Rhinoceros Party has decided on the following strategy. Every day all dole applicants will be placed in a large circle, facing inwards. Someone is arbitrarily chosen as number 1, and the rest are numbered counterclockwise up to N (who will be standing on 1’s left). Starting from 1 and moving counter-clockwise, one labour official counts off k applicants, while another official starts from N and moves clockwise, counting m applicants. The two who are chosen are then sent off for retraining; if both officials pick the same person she (he) is sent off to become a politician. Each official then starts counting again at the next available person and the process continues until no-one is left. Note that the two victims (sorry, trainees) leave the ring simultaneously, so it is possible for one official to count a person already selected by the other official.Input
Writeaprogramthatwillsuccessivelyreadin(inthatorder)thethreenumbers(N, k and m; k,m > 0, 0 < N < 20) and determine the order in which the applicants are sent off for retraining. Each set of three numbers will be on a separate line and the end of data will be signalled by three zeroes (0 0 0).Output
For each triplet, output a single line of numbers specifying the order in which people are chosen. Each number should be in a field of 3 characters. For pairs of numbers list the person chosen by the counterclockwise official first. Separate successive pairs (or singletons) by commas (but there should not be a trailing comma).Note: The symbol ⊔ in the Sample Output below represents a space.
Sample Input
10 4 3 0 0 0Sample Output
␣␣4␣␣8,␣␣9␣␣5,␣␣3␣␣1,␣␣2␣␣6,␣10,␣␣7解题心得:
紫书上的例题,很简单可以直接模拟整个过程就行了,但是要注意输出的问题(”%3d”)。在一个功能的模块需要多次使用的时候可以直接写成函数,减少代码长度,不容易出错。
#include<bits/stdc++.h> using namespace std; const int maxn = 1e5; bool cir[maxn]; int r,l,Len; int _find(int pos,int len,int dir)//dir代表方向,1代表向右,0代表向左,pos是当前位置,len代表要走多少步 { while(len) { if(dir == 1) pos++; else pos--; if(pos > Len) pos = 1; if(pos < 1) pos = Len; if(!cir[pos]) len--; } return pos; } int main() { int len,m,k; while(scanf("%d%d%d",&len,&m,&k) && len+m+k) { Len = len; memset(cir,false,sizeof(cir)); bool flag = false; r = 0,l = Len+1; while(len) { r = _find(r,m,1); l = _find(l,k,0); cir[r] = cir[l] = true;//true代表已经退出了的 if(r == l) { if(!flag)//flag用来控制,的输出 printf("%3d",r); else printf(",%3d",r); len--; } else { if(!flag) printf("%3d%3d",r,l); else printf(",%3d%3d",r,l); len -= 2; } flag = true; } printf("\n"); } return 0; }
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